Problem 75 - Precalculus Student Solutions Manual 5th

Question

For each function as defined that is one-to-one, (a) write an equation for the inverse function in the form y = ƒ -11x2, (b) graph ƒ and ƒ -1 on the same axes, and (c) give the domain and the range of ƒ and ƒ -1. If the function is not one-to-one, say so. $$f(x)=\sqrt{6+x}, \quad x \geq-6$$

Step-by-Step Solution

Answer

Inverse function: \( f^{-1}(x) = x^2 - 6 \), domain and range of \( f \): \([-6, \infty) \) and \([0, \infty) \), domain and range of \( f^{-1} \): \([0, \infty) \) and \([-6, \infty) \).

Step 1: Verify One-to-One Property

To verify if the function is one-to-one, check if every y-value has a unique x-value. For the function \( f(x) = \sqrt{6+x} \) with \( x \geq -6 \), as \( x \) increases, \( f(x) \) increases. Therefore, every \( y \) has a unique corresponding \( x \). Thus, the function is one-to-one.

Step 2: Find the Inverse Function

To find the inverse, solve \( y = \sqrt{6 + x} \) for \( x \). Start by squaring both sides: \( y^2 = 6 + x \). Then isolate \( x \): \( x = y^2 - 6 \). Rewrite the equation in inverse function form: \( f^{-1}(x) = x^2 - 6 \).

Step 3: Graph \( f \) and \( f^{-1} \)

Graph \( f(x) = \sqrt{6 + x} \) and its inverse \( f^{-1}(x) = x^2 - 6 \) on the same axes. The graph of \( f(x) \) is a square root curve starting at (-6, 0) and increasing. The graph of \( f^{-1}(x) \) is a parabola opening upwards with vertex at (0, -6). Both graphs will be symmetrical about the line \( y=x \).

Step 4: Determine Domain and Range

The domain of \( f(x) \) is the set of all \( x \) for which the function is defined: \( x \geq -6 \). The range of \( f(x) \) is all possible output values: \( y \geq 0 \). For the inverse function \( f^{-1}(x) \), the domain is the range of the original function: \( x \geq 0 \). The range of \( f^{-1}(x) \) is the domain of the original function: \( y \geq -6 \).

Key Concepts

One-to-One Functions

A function is called one-to-one if each value in the range is mapped from a unique value in the domain.
This means no two different domain values map to the same range value.
In simpler terms, each output value has one unique input value.
For the function given in our problem, \(f(x) = \sqrt{6+x}\), we checked that as x increases, the y-value (output) also increases.
There are no repeated y-values for different x-values, confirming that the function is one-to-one.
This property is crucial because only one-to-one functions have inverses that are also functions.

Domain and Range

The domain and range of a function describe the set of possible input and output values.
For the function \(f(x) = \sqrt{6 + x}\) defined as \ x \geq -6 \, the domain is all real numbers greater than or equal to -6.
This is because the expression inside the square root, \6 + x\, must be non-negative.
The range is all outputs the function can generate, \ y \geq 0\, since square roots are non-negative.
For the inverse function \ y = f^{-1}(x) = x^2 - 6 \, the domain is \ x \geq 0 \, and the range is \ y \geq -6 \,.
This relationship between the domain and range helps in understanding how the functions behave and graphically intersect.

Graphing Functions

Graphing functions is an essential skill in understanding their behavior visually.
To graph \(f(x) = \sqrt{6 + x}\), you start at the point (-6, 0) and plot a curve that increases steadily.
This type of curve is typical for square root functions.
For its inverse, \ y = x^2 - 6 \, you plot a parabola starting from (0, -6) that opens upwards.
The graph of a function and its inverse are mirror images about the line y = x.
This symmetry helps to verify the correctness of the inverse function.
By plotting both functions, you can visually see their possible input and output values, thus reinforcing the concept of domain and range.